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### Prove or Disprove that Sqrt(2) and Sqrt(4) are Irrational

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Date: 23 Jun 1995 08:21:58 -0400
From: Mike Rubitschka
Subject: Problems with a mathematics exercise

I have learned the following proof that sqrt(2) is irrational:

Suppose for contradiction that sqrt(2) is rational.
Then sqrt(2) = m/n
Assume that m and n are not both even.
sqrt(2) = m/n ; m = Sqrt(2)*n and hence
m*m = 2*n*n
This shows that m*m is an even number. Thus, m itself is even.
This means m = 2*p for some integer p.
But then 4*p*p = m*m = 2*n*n
or       n*n = 2*p*p
This shows that n*n is even, and so it follows that n is even.
But that contradicts the fact that not both m and n are even ...
So sqrt(2) must be irrational.

Problem:
If you use this proof with sqrt(4) you can proof that sqrt(4) is
irrational!!! :

Suppose for contradiction that sqrt(4) is rational.
Then sqrt(4) = m/n
Assume that m and n are not both divisible by 4.
sqrt(4) = m/n ; m = Sqrt(4)*n and hence
m*m = 4*n*n
This shows that m*m is divisible by 4. Thus, m itself is div. by 4.
This means m = 4*p for some integer p.
But then 16*p*p = m*m = 4*n*n
or        n*n = 4*p*p
This shows that n*n is div by 4, and so it follows that n is div. by 4.
But that contradicts the fact that not both m and n are div. by 4 ...
So sqrt(4) must be irrational???

Where is the fault which I must have done?

Michael Rubitschka
Vienna, Austria
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Date: 23 Jun 1995 09:34:59 -0400
From: Dr. Ethan
Subject: Re: Problems with a mathematics exercise

Hey Mike,

Well, That is a great question.  and if your method was right we
would have quite a problem. But here is the problem.  You say:

" This shows that m*m is divisible by 4. Thus, m itself is div. by 4."

Well that isn't true.  What if m is 2 then 2*2 is divisible by 4 but 2 isn't.

Does that make sense?

Ethan - Doctor On Call
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Associated Topics:
High School Exponents

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